10. Newton’s second law - Revision history

Ian at 13:06, 27 April 2011

2011-04-27T13:06:53Z

←Older revision		Revision as of 13:06, 27 April 2011
Line 2:		Line 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}	+	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}
	{{Not selected tab\|[[10. Exercises\|Exercises]]}}		{{Not selected tab\|[[10. Exercises\|Exercises]]}}
		+	{{Not selected tab\|[[10. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
-
-

Ian at 10:01, 6 March 2011

2011-03-06T10:01:35Z

←Older revision		Revision as of 10:01, 6 March 2011
Line 210:		Line 210:
	<math>\begin{align}		<math>\begin{align}
	& 20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ =3a \\		& 20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ =3a \\
-	& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4\textrm{.}48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\	+	& a=\frac{20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ }{3}=4\textrm{.}48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Ian at 12:23, 14 June 2010

2010-06-14T12:23:43Z

←Older revision		Revision as of 12:23, 14 June 2010
Line 27:		Line 27:

	<math>F = \mu R</math>		<math>F = \mu R</math>
		+
		+	Newton's Second Law is sometimes referred to as the Law of Motion.

Ian at 13:32, 23 May 2010

2010-05-23T13:32:06Z

←Older revision		Revision as of 13:32, 23 May 2010
Line 23:		Line 23:

	<math>\mathbf{F}=m\mathbf{a}</math>		<math>\mathbf{F}=m\mathbf{a}</math>
		+
		+	If there is friction the friction equation holds.
		+
		+	<math>F = \mu R</math>

Ian at 08:53, 20 May 2010

2010-05-20T08:53:51Z

←Older revision		Revision as of 08:53, 20 May 2010
Line 17:		Line 17:

	<math>F=ma</math>		<math>F=ma</math>
		+
		+	If the acceleration is not known one assumes a direction for the acceleration <math>a</math> and applies this law in that direction.

	The vector form of this equation is		The vector form of this equation is

Ian at 13:35, 8 March 2010

2010-03-08T13:35:47Z

←Older revision		Revision as of 13:35, 8 March 2010
Line 14:		Line 14:
	Newton's Second Law		Newton's Second Law

-	If the resultant force on a particle is not zero, then it will ~~accelerate.~~	+	If the resultant force <math>F</math> on a particle of mass <math>m</math> is not zero, then it will have an acceleration <math>a</math> given by,

	<math>F=ma</math>		<math>F=ma</math>
		+
		+	The vector form of this equation is

	<math>\mathbf{F}=m\mathbf{a}</math>		<math>\mathbf{F}=m\mathbf{a}</math>

Ian at 18:53, 16 February 2010

2010-02-16T18:53:40Z

←Older revision		Revision as of 18:53, 16 February 2010
Line 83:		Line 83:
	First find the resultant force acting on the car.		First find the resultant force acting on the car.

-	<math>~~\text{Resultant Force }~~=\text{ }3000-R</math>	+	Resultant Force <math>=\text{ }3000-R</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Line 110:		Line 110:
	First find the resultant force acting on the car.		First find the resultant force acting on the car.

-	<math>~~\text{Resultant Force }~~=\text{ }5000-2000=3000\text{ N}</math>	+	Resultant Force <math>=\text{ }5000-2000=3000\text{ N}</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Line 142:		Line 142:
	The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.		The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.

-	<math>~~\text{Resultant Force }~~=11760\sin 3{}^\circ -400</math>	+	Resultant Force <math>=11760\sin 3{}^\circ -400</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Line 186:		Line 186:
	Now resolving horizontally, the resultant force can be found.		Now resolving horizontally, the resultant force can be found.

		+	Resultant Force
		+
	<math>\begin{align}		<math>\begin{align}
-	& ~~\text{Resultant Force }~~=20\cos 30{}^\circ -F \\	+	& =20\cos 30{}^\circ -F \\
	& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\		& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\
	& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ		& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ

Ian at 18:47, 16 February 2010

2010-02-16T18:47:44Z

←Older revision		Revision as of 18:47, 16 February 2010
Line 56:		Line 56:

	a) If		a) If
-	<math>a=0.2</math>	+	<math>a=0\textrm{.}2</math>
	, then		, then
-	<math>T=2940+300\times 0.2=3000\text{ N}</math>	+	<math>T=2940+300\times 0\textrm{.}2=3000\text{ N}</math>
	.		.

	b) If		b) If
-	<math>a=-0.05</math>	+	<math>a=-0\textrm{.}05</math>
	(since the lift is accelerating downwards then:		(since the lift is accelerating downwards then:

-	<math>T=2940+300\times (-0.05)=2925\text{ N}</math>	+	<math>T=2940+300\times (-0\textrm{.}05)=2925\text{ N}</math>
	.		.

Line 90:		Line 90:

	<math>\begin{align}		<math>\begin{align}
-	& 3000-R=1200\times 1.8 \\	+	& 3000-R=1200\times 1\textrm{.}8 \\
	& R=3000-2160 \\		& R=3000-2160 \\
	& =840\text{ N}		& =840\text{ N}
Line 118:		Line 118:
	<math>\begin{align}		<math>\begin{align}
	& 3000=2500a \\		& 3000=2500a \\
-	& a=\frac{3000}{2500}=1.2\text{ m}{{\text{s}}^{\text{-2}}}	+	& a=\frac{3000}{2500}=1\textrm{.}2\text{ m}{{\text{s}}^{\text{-2}}}
	\end{align}</math>		\end{align}</math>

Line 150:		Line 150:
	<math>\begin{align}		<math>\begin{align}
	& 11760\sin 3{}^\circ -400=1200a \\		& 11760\sin 3{}^\circ -400=1200a \\
-	& a=\frac{11760\sin 3{}^\circ -400}{1200}=0.180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\	+	& a=\frac{11760\sin 3{}^\circ -400}{1200}=0\textrm{.}180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Line 171:		Line 171:

	<math>\begin{align}		<math>\begin{align}
-	& R+20\sin 30{}^\circ =29.4 \\	+	& R+20\sin 30{}^\circ =29\textrm{.}4 \\
-	& R=29.4-20\sin 30{}^\circ \\	+	& R=29\textrm{.}4-20\sin 30{}^\circ \\
	\end{align}</math>		\end{align}</math>

Line 180:		Line 180:

	<math>\begin{align}		<math>\begin{align}
-	& F=0.2\times \left( 29.4-20\sin 30{}^\circ \right) \\	+	& F=0.2\times \left( 29\textrm{.}4-20\sin 30{}^\circ \right) \\
-	& =5.88-4\sin 30{}^\circ	+	& =5\textrm{.}88-4\sin 30{}^\circ
	\end{align}</math>		\end{align}</math>

Line 188:		Line 188:
	<math>\begin{align}		<math>\begin{align}
	& \text{Resultant Force }=20\cos 30{}^\circ -F \\		& \text{Resultant Force }=20\cos 30{}^\circ -F \\
-	& =20\cos 30{}^\circ -(5.88-4\sin 30{}^\circ ) \\	+	& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\
-	& =20\cos 30{}^\circ -5.88+4\sin 30{}^\circ	+	& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ
	\end{align}</math>		\end{align}</math>

Line 197:		Line 197:

	<math>\begin{align}		<math>\begin{align}
-	& 20\cos 30{}^\circ -5.88+4\sin 30{}^\circ =3a \\	+	& 20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ =3a \\
-	& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4.48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\	+	& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4\textrm{.}48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Ian at 17:46, 16 February 2010

2010-02-16T17:46:51Z

←Older revision		Revision as of 17:46, 16 February 2010
Line 46:		Line 46:
	Consider the resultant force on the particle, taking the upward direction as positive.		Consider the resultant force on the particle, taking the upward direction as positive.

-	<math>~~\text{Resultant Force }~~=T-2940</math>	+	Resultant Force <math>=T-2940</math>

	Applying Newton’s second law gives:		Applying Newton’s second law gives:
Line 156:		Line 156:
	'''[[Example 10.5]]'''		'''[[Example 10.5]]'''

-	A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at ~~30~~ to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.	+	A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at <math>30
		+	^\circ</math> to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.

	[[Image:TF10.5.GIF]]		[[Image:TF10.5.GIF]]

Ian at 17:42, 16 February 2010

2010-02-16T17:42:50Z

(Difference between revisions)

←Older revision		Revision as of 13:06, 27 April 2011
Line 2:		Line 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}	+	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}
	{{Not selected tab\|[[10. Exercises\|Exercises]]}}		{{Not selected tab\|[[10. Exercises\|Exercises]]}}
		+	{{Not selected tab\|[[10. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
-
-